Problem: Let $f(x)=-x^4+8x^2-8$. What is the absolute minimum value of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-60$ (Choice B) B $-12$ (Choice C) C $-8$ (Choice D) D $f$ has no minimum value
Let's first find the relative extremum points of $f$, and then consider them along with the function's end behavior in both directions. We start with finding the critical points of $f$. The derivative of $f$ is $f'(x)=-4x(x+2)(x-2)$. $f'(x)=0$ for $x=-2,0,2$. $f'$ is defined for all real numbers. Therefore, our critical points are $x=-2$, $x=0$, and $x=2$. Our critical points divide the function's domain (which is all real numbers) into four intervals: $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $(-\infty, \llap{-}2)$ $( \llap{-}2,0)$ $(0,2)$ $(2,\infty)$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $(-\infty,-2)$ $x=-3$ $f'(-3)=60>0$ $f$ is increasing $\nearrow$ $(-2,0)$ $x=-1$ $f'(-1)=-12<0$ $f$ is decreasing $\searrow$ $(0,2)$ $x=1$ $f'\left(1\right)=12>0$ $f$ is increasing $\nearrow$ $(2,\infty)$ $x=3$ $f'(3)=-60<0$ $f$ is decreasing $\searrow$ Now let's look at all the critical points: $x$ $f(x)$ Before After Verdict $-2$ $8$ $\nearrow$ $\searrow$ Maximum $0$ $-8$ $\searrow$ $\nearrow$ Minimum $2$ $8$ $\nearrow$ $\searrow$ Maximum Let's imagine ourselves walking on the graph of $f$, starting all the way to the left (from $-\infty$ ) and going all the way to the right (until $+\infty$ ). According to the table, we will start by going up until $(-2,8)$, then go down until $(0,-8)$, then up again until $(2,8)$, and then forever go down. This means that $\lim_{x\to-\infty}f(x)=\lim_{x\to +\infty}f(x)=-\infty$, which means $f$ has no minimum value. In conclusion, $f$ has no absolute minimum value.